Question

In figure, tangent segments PS and PT are drawn to a circle with centre O such that ∠SPT = 120°. Prove that OP = 2 PS.

Explain with complete steps & reasoning.



PS : Can this question be solved with trigonometric ratios? If yes, do so.

Answer 1

Heya,

This, question can be solved by trigonometric ratios.

Given : PT and PS are tangent of the circle.
O is the centre of the circle and ∠SPT = 120°

To prove : OP = 2 PS.

Proof : In ΔOPS and ΔOPT

∠OSP = ∠OTP (each 90° bcz radius is perpendicular to the tangent)

PS = PT (tangent from an external point are equal)

OS = OT (radius of the circle are equal)

So,

ΔOPS ≅ ΔOPT (by SAS similarity criterion)

Therefore,

∠OPS = ∠OPT (by CPCT)
∠POS = ∠POT (by CPCT)

As,
∠SPT = 120° and ∠SOT = 60°
So, now;

∠OPS = ∠OPT = 60°
∠POS = ∠POT = 30°

Now, by taking ΔPOS

We get,

=> sin30° = PS/PO

=> 1/2 = PS/PO

=> PO = 2PS

Hence Proved.

Hope this helps....:)

If this proving is not understood by u then u can re-ask me I will let u understand.

Answer 2

Hi ,

Given : PS , PT are tangent segments drawn

to a circle with center O.

< SPT = 120°.

RTP : OP = 2PS

proof :

In ∆OSP and ∆OTP

OS = OT ( radii of a same circle )

OP = OP ( common side )

PS = PT ( lengths of tangents from an external

point are equal )

Therefore ,

∆OSP = ∆OTP [ SSS congruence rule ]

<SPO = <OPT = 120/2 = 60° [ CPCT ]

Now ,

In ∆ OSP ,

<SOP = 30°

In ∆ OSP rt .angled Triangle ,

sin 30° = PS/OP

1/2 = PS/OP

OP = 2PS

Hence proved.

I hope this helps you.

: )